2a. A wave has a frequency of 50 Hz and a wavelength of 0.10 m. How do you find the wave speed? *

A cat runs and jumps from one roof top to another which is 5 m away and 3 m below. Calculate the minimum horizontal speed with w

icang [17]

ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s

6 0

3 years ago

the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes

slamgirl [31]

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.

5 0

3 years ago

A space probe produces a radio signal pulse. If the pulse reaches Earth 12.3 seconds after it is emitted by the probe, what is t

adoni [48]

Answer:

Distance = 3.69 × 10^9 m

The distance from the probe to Earth is 3.69 × 10^9 m

Explanation:

Distance from the probe to the Earth can be derived using the simple motion formula;

Distance = speed × time .....1

Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.

Speed of radio signal = speed of light = 3.0 × 10^8 m/s

time taken to reach the earth = 12.3 seconds

Substituting the values of speed and time into equation 1;

Distance = 3.0 × 10^8 m/s × 12.3 s

Distance = 36.9 × 10^8 m

Distance = 3.69 × 10^9 m

Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s

5 0

3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write: